$\int t^4\,dt=$ $+C$
The integrand is of the form $x^n$ where $n\neq-1$, so we can use the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ $\begin{aligned} \int t^{{4}}\,dt&=\dfrac{t^{{4}+1}}{{4}+1}+C \\\\ &=\dfrac15 t^5+C \end{aligned}$ In conclusion, $\int t^4\,dt=\dfrac15 t^5+C$